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Interior, exterior, and boundary of deleted neighborhood
Limit points and interior pointsInterior, Exterior Boundary of a subset with irrational constraintsClosure, interior and boundary pointFor a set E, define interior, exterior, and boundary pointsWhy must an interior point of $E$ be an element of $E$?Determine the interior, boundary, exterior and closure of the set $S= (x_1,…,x_n)inmathbb R^nmid forall x_iin mathbb Q$Find the boundary, the interior and exterior of a set.How do I prove that the boundary of a set has the property that an open ball around it will have at least one interior and one exterior point?Topology: interior points and boundary pointsAny open neighborhood of a boundary point contains an interior point and an exterior point.
$begingroup$
Let the set A be a punctured disk around p = (0,0) of radius 1 in R^2. Is the point p part of the interior, exterior, or boundary of A?
Certainly p is a limit point of A. As a limit point of A, it is not in the exterior of A.
Now an open disk around p of radius > 0 is not contained in A because p is not contained in A.
But also a deleted neighborhood around p of radius > 0 does not contain points in both A and A^c.
So the exterior is out of the question, but I can't seem to place p in the interior or in the boundary. Interior is defined with a neighborhood and boundary is defined with a deleted neighborhood and that is what seems to be the issue. Surely p must be in either the interior, exterior, or boundary.
real-analysis general-topology metric-spaces
asked Mar 10 at 20:53
Dylan MehrerDylan Mehrer
31
New contributor
Dylan Mehrer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let the set A be a punctured disk around p = (0,0) of radius 1 in R^2. Is the point p part of the interior, exterior, or boundary of A?
Certainly p is a limit point of A. As a limit point of A, it is not in the exterior of A.
Now an open disk around p of radius > 0 is not contained in A because p is not contained in A.
But also a deleted neighborhood around p of radius > 0 does not contain points in both A and A^c.
So the exterior is out of the question, but I can't seem to place p in the interior or in the boundary. Interior is defined with a neighborhood and boundary is defined with a deleted neighborhood and that is what seems to be the issue. Surely p must be in either the interior, exterior, or boundary.
real-analysis general-topology metric-spaces
asked Mar 10 at 20:53
Dylan MehrerDylan Mehrer
31
New contributor
Dylan Mehrer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
What is your definition of the boundary of a set? Mine is that the boundary of $S$, denoted $partial S$, is $barSsetminus S^circ$ (that is, the closure minus the interior).
$endgroup$
– ItsJustVennDiagramsBro
Mar 10 at 20:56
$begingroup$
What's a 'deleted' neighborhood? Notice that in any neighborhood of $p$ there's $p$ itself that is in $A^c$
$endgroup$
– Exodd
Mar 10 at 21:01
$begingroup$
No definition of boundary was given other than the following theorem: If X has non-trivial metric d and A is a subset of X then p is a boundary point of A iff for any r > 0 the ball around p of radius r contains at least one point in A and at least one point in A^c other than p itself
$endgroup$
– Dylan Mehrer
Mar 10 at 21:25
$begingroup$
Deleted neighborhood is that "other than p itself" part
$endgroup$
– Dylan Mehrer
Mar 10 at 21:28
add a comment |
$begingroup$
Let the set A be a punctured disk around p = (0,0) of radius 1 in R^2. Is the point p part of the interior, exterior, or boundary of A?
Certainly p is a limit point of A. As a limit point of A, it is not in the exterior of A.
Now an open disk around p of radius > 0 is not contained in A because p is not contained in A.
But also a deleted neighborhood around p of radius > 0 does not contain points in both A and A^c.
So the exterior is out of the question, but I can't seem to place p in the interior or in the boundary. Interior is defined with a neighborhood and boundary is defined with a deleted neighborhood and that is what seems to be the issue. Surely p must be in either the interior, exterior, or boundary.
real-analysis general-topology metric-spaces
asked Mar 10 at 20:53
Dylan MehrerDylan Mehrer
31
New contributor
Dylan Mehrer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let the set A be a punctured disk around p = (0,0) of radius 1 in R^2. Is the point p part of the interior, exterior, or boundary of A?
Certainly p is a limit point of A. As a limit point of A, it is not in the exterior of A.
Now an open disk around p of radius > 0 is not contained in A because p is not contained in A.
But also a deleted neighborhood around p of radius > 0 does not contain points in both A and A^c.
So the exterior is out of the question, but I can't seem to place p in the interior or in the boundary. Interior is defined with a neighborhood and boundary is defined with a deleted neighborhood and that is what seems to be the issue. Surely p must be in either the interior, exterior, or boundary.
real-analysis general-topology metric-spaces
real-analysis general-topology metric-spaces
asked Mar 10 at 20:53
Dylan MehrerDylan Mehrer
31
New contributor
Dylan Mehrer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 20:53
Dylan MehrerDylan Mehrer
31
New contributor
Dylan Mehrer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 20:53
Dylan MehrerDylan Mehrer
31
New contributor
Dylan Mehrer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 20:53
Dylan MehrerDylan Mehrer
31
asked Mar 10 at 20:53
Dylan MehrerDylan Mehrer
31
31
New contributor
Dylan Mehrer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dylan Mehrer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dylan Mehrer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
What is your definition of the boundary of a set? Mine is that the boundary of $S$, denoted $partial S$, is $barSsetminus S^circ$ (that is, the closure minus the interior).
$endgroup$
– ItsJustVennDiagramsBro
Mar 10 at 20:56
$begingroup$
What's a 'deleted' neighborhood? Notice that in any neighborhood of $p$ there's $p$ itself that is in $A^c$
$endgroup$
– Exodd
Mar 10 at 21:01
$begingroup$
No definition of boundary was given other than the following theorem: If X has non-trivial metric d and A is a subset of X then p is a boundary point of A iff for any r > 0 the ball around p of radius r contains at least one point in A and at least one point in A^c other than p itself
$endgroup$
– Dylan Mehrer
Mar 10 at 21:25
$begingroup$
Deleted neighborhood is that "other than p itself" part
$endgroup$
– Dylan Mehrer
Mar 10 at 21:28
add a comment |
$begingroup$
What is your definition of the boundary of a set? Mine is that the boundary of $S$, denoted $partial S$, is $barSsetminus S^circ$ (that is, the closure minus the interior).
$endgroup$
– ItsJustVennDiagramsBro
Mar 10 at 20:56
$begingroup$
What's a 'deleted' neighborhood? Notice that in any neighborhood of $p$ there's $p$ itself that is in $A^c$
$endgroup$
– Exodd
Mar 10 at 21:01
$begingroup$
No definition of boundary was given other than the following theorem: If X has non-trivial metric d and A is a subset of X then p is a boundary point of A iff for any r > 0 the ball around p of radius r contains at least one point in A and at least one point in A^c other than p itself
$endgroup$
– Dylan Mehrer
Mar 10 at 21:25
$begingroup$
Deleted neighborhood is that "other than p itself" part
$endgroup$
– Dylan Mehrer
Mar 10 at 21:28
$begingroup$
What is your definition of the boundary of a set? Mine is that the boundary of $S$, denoted $partial S$, is $barSsetminus S^circ$ (that is, the closure minus the interior).
$endgroup$
– ItsJustVennDiagramsBro
Mar 10 at 20:56
$begingroup$
What is your definition of the boundary of a set? Mine is that the boundary of $S$, denoted $partial S$, is $barSsetminus S^circ$ (that is, the closure minus the interior).
$endgroup$
– ItsJustVennDiagramsBro
Mar 10 at 20:56
$begingroup$
What's a 'deleted' neighborhood? Notice that in any neighborhood of $p$ there's $p$ itself that is in $A^c$
$endgroup$
– Exodd
Mar 10 at 21:01
$begingroup$
What's a 'deleted' neighborhood? Notice that in any neighborhood of $p$ there's $p$ itself that is in $A^c$
$endgroup$
– Exodd
Mar 10 at 21:01
$begingroup$
No definition of boundary was given other than the following theorem: If X has non-trivial metric d and A is a subset of X then p is a boundary point of A iff for any r > 0 the ball around p of radius r contains at least one point in A and at least one point in A^c other than p itself
$endgroup$
– Dylan Mehrer
Mar 10 at 21:25
$begingroup$
No definition of boundary was given other than the following theorem: If X has non-trivial metric d and A is a subset of X then p is a boundary point of A iff for any r > 0 the ball around p of radius r contains at least one point in A and at least one point in A^c other than p itself
$endgroup$
– Dylan Mehrer
Mar 10 at 21:25
$begingroup$
Deleted neighborhood is that "other than p itself" part
$endgroup$
– Dylan Mehrer
Mar 10 at 21:28
$begingroup$
Deleted neighborhood is that "other than p itself" part
$endgroup$
– Dylan Mehrer
Mar 10 at 21:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$mathrmInt(A) subset A$ so $p notin mathrmInt(A)$.
But $p$ is in the closure $overlineA$ of $A$, so $p in overlineA setminus mathrmInt(A)$, i.e. $p$ is in the boundary of $A$.
answered Mar 10 at 21:03
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
$begingroup$
Ok but the question before this one asked to show that p is a limit point, but not a boundary point.
$endgroup$
– Dylan Mehrer
Mar 10 at 21:24
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$mathrmInt(A) subset A$ so $p notin mathrmInt(A)$.
But $p$ is in the closure $overlineA$ of $A$, so $p in overlineA setminus mathrmInt(A)$, i.e. $p$ is in the boundary of $A$.
answered Mar 10 at 21:03
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
$begingroup$
Ok but the question before this one asked to show that p is a limit point, but not a boundary point.
$endgroup$
– Dylan Mehrer
Mar 10 at 21:24
add a comment |
$begingroup$
$mathrmInt(A) subset A$ so $p notin mathrmInt(A)$.
But $p$ is in the closure $overlineA$ of $A$, so $p in overlineA setminus mathrmInt(A)$, i.e. $p$ is in the boundary of $A$.
answered Mar 10 at 21:03
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
$begingroup$
Ok but the question before this one asked to show that p is a limit point, but not a boundary point.
$endgroup$
– Dylan Mehrer
Mar 10 at 21:24
add a comment |
$begingroup$
$mathrmInt(A) subset A$ so $p notin mathrmInt(A)$.
But $p$ is in the closure $overlineA$ of $A$, so $p in overlineA setminus mathrmInt(A)$, i.e. $p$ is in the boundary of $A$.
answered Mar 10 at 21:03
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
$mathrmInt(A) subset A$ so $p notin mathrmInt(A)$.
But $p$ is in the closure $overlineA$ of $A$, so $p in overlineA setminus mathrmInt(A)$, i.e. $p$ is in the boundary of $A$.
answered Mar 10 at 21:03
TheSilverDoeTheSilverDoe
3,837112
answered Mar 10 at 21:03
TheSilverDoeTheSilverDoe
3,837112
answered Mar 10 at 21:03
TheSilverDoeTheSilverDoe
3,837112
answered Mar 10 at 21:03
TheSilverDoeTheSilverDoe
3,837112
3,837112
$begingroup$
Ok but the question before this one asked to show that p is a limit point, but not a boundary point.
$endgroup$
– Dylan Mehrer
Mar 10 at 21:24
add a comment |
$begingroup$
Ok but the question before this one asked to show that p is a limit point, but not a boundary point.
$endgroup$
– Dylan Mehrer
Mar 10 at 21:24
$begingroup$
Ok but the question before this one asked to show that p is a limit point, but not a boundary point.
$endgroup$
– Dylan Mehrer
Mar 10 at 21:24
$begingroup$
Ok but the question before this one asked to show that p is a limit point, but not a boundary point.
$endgroup$
– Dylan Mehrer
Mar 10 at 21:24
add a comment |
Dylan Mehrer is a new contributor. Be nice, and check out our Code of Conduct.
Dylan Mehrer is a new contributor. Be nice, and check out our Code of Conduct.
Dylan Mehrer is a new contributor. Be nice, and check out our Code of Conduct.
Dylan Mehrer is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
What is your definition of the boundary of a set? Mine is that the boundary of $S$, denoted $partial S$, is $barSsetminus S^circ$ (that is, the closure minus the interior).
$endgroup$
– ItsJustVennDiagramsBro
Mar 10 at 20:56
$begingroup$
What's a 'deleted' neighborhood? Notice that in any neighborhood of $p$ there's $p$ itself that is in $A^c$
$endgroup$
– Exodd
Mar 10 at 21:01
$begingroup$
No definition of boundary was given other than the following theorem: If X has non-trivial metric d and A is a subset of X then p is a boundary point of A iff for any r > 0 the ball around p of radius r contains at least one point in A and at least one point in A^c other than p itself
$endgroup$
– Dylan Mehrer
Mar 10 at 21:25
$begingroup$
Deleted neighborhood is that "other than p itself" part
$endgroup$
– Dylan Mehrer
Mar 10 at 21:28